Lecture 13, p 1

“But why must I treat the measuring device classically? What will happen to me if I don’t??”

--Eugene Wigner“There is obviously no such limitation – I can measure the energy and look at my watch; then I know both energy and time!” --L. D. Landau, on the time-energy uncertainty principle

Lecture 13, p 2

Lab 3 Comments

Lab 3 meets this week if you are normally in 132 Loomis.Lab 3 meets next week if you are normally in 164 Loomis.So does Discussion, and there is a quiz, so don’t skip...

For the lab:You will need your “Active Directory” Login

See: http://www.ad.uiuc.edu

You can save a lot of time by reading the lab ahead of time. It’s a tutorial on how to draw wave functions

Lecture 13, p 3

Lecture 13: Superposition &Time-Dependent Quantum

States

x

|(x,t0)|2U= U=

0 xL

|(x,t=0)|2U= U=

0 xL

Lecture 13, p 4

Last WeekTime-independent Schrodinger’s Equation (SEQ):

It describes a particle that has a definite energy, E. The solutions, (x), are time independent (stationary states).

We considered two potentials, U(x):

Finite-depth square well Boundary conditions. Particle can “leak” into forbidden region. Comparison with infinite-depth well.

Harmonic oscillator Energy levels are equally spaced. A good approximation in many problems.

)()()()(2 2

22

xExxUdxxd

m

Lecture 13, p 5

Today

Time dependent SEQ: Superposition of states and particle motion

Measurement in quantum physics

Time-energy uncertainty principle

Lecture 13, p 6

Time-Dependent SEQTo explore how particle wave functions evolve with time, which is useful for a number of applications as we shall see, we need to consider the time-dependent SEQ:

Changes from the time independent version: E iħ d/dt We no longer assume a definite E. (x) (x,t) The solutions will have time dependence. i = (-1) appears The solutions will be complex.

This equation describes the complete time and space dependence of a quantum particle in a potential U(x).It replaces the classical particle dynamics law, F=ma.

The SEQ is linear in , and so the Superposition Principle applies:If 1 and 2 are solutions to the time-dependent SEQ, then so is any linear combination of 1 and 2 (example: = 0.61 + 0.8i2)

2 2

2

( , ) ( , )( ) ( , )2d x t d x tU x x t i

m dx dt

Lecture 13, p 7

Review of Complex NumbersThe equation, ei = cos + isin, might be new to you. It is a convenient way to represent complex numbers. It also (once you are used to it) makes trigonometry simpler.

a. Draw an Argand diagram of ei.

The Argand diagram of a complex number, A, puts Re(A) on the x-axis and Im(A) on the y-axis. Notice the trig relation between the x and y components. is the angle of A from the real axis. In an Argand diagram, ei looks like a vector of length 1, and components (cos, sin).Re(A)

Im(A)

A = ei

The quantity, cei (c and both real), is a complex number of magnitude |c|. The magnitude of a complex number, A, is |A| = (A*A), where A* is the complex conjugate of A.

Re(A)

Im(A)

= t

A = eit

b. Suppose that varies with time, = t. How does the Argand diagram behave?

At t = 0, = 0, so A = 1 (no imaginary component). As time progresses, A rotates counterclockwise with angular frequency . This is the math that underlies phasors.

Lecture 13, p 8

Act i

1. What is (-i)i ?

a. –i b. -1 c. +1

2. What is 1/i?

a. –1 b. -i c. +i

3. What is |ei|2 ?

a. 0 b. e2i c. 1

1i

Lecture 13, p 9

Solution

1. What is (-i)i ?

a. –i b. -1 c. +1

2. What is 1/i?

a. –1 b. -i c. +i

3. What is |ei|2 ?

a. 0 b. e2i c. 1

1i

2( ) ( 1) 1 i i i

1 11

i i i

i i i

*2

0 1

i i i i i

i i

e e e e e

e eThis is the definition of the magnitude of a complex number.

Lecture 13, p 10

Time-dependence of Energy EigenstatesThe time-independent SEQ is just a special case of the time-dependent SEQ.So, if (x,t) is a state with definite energy, it is a solution to both equations.Both equations have the same left-hand side, so the right sides must be equal:

This equation has the solution:

is complex. However, we are interested in ||2, because that’s what we measure.

So, for an energy eigenstate:the probability density, ||2, has no time dependence! (i.e., it’s a “stationary state”.)

We don’t actually need the time-independent SEQ, but if we know we’re dealing with energy eigenstates, the math is simpler.

( , ) h it( w) i t Ex t x e

2 2*( , ) ( ) ( ) ( )i t i tx t x e x e x

( , )( , ) d x tE x t idt

RHS of time ind. SEQ

RHS of time dep. SEQ

Notes: (x) is not determined.

We need the LHS for that. E = ħ = hf, as expected.

This is always a real number.

Lecture 13, p 11

Example: Motion of a Free ParticleA free particle moves without applied forces; so set U(x) = 0.The SEQ is now reasonably simple:

The second x-derivative is proportional to the first t-derivative.Here’s one solution:

This is a traveling wave. The particle hasMomentum: p = ħk = h/Energy: E = ħ = hf .

Not a surprise, I hope. 2 2

2

( , ) ( , )2d x t d x ti

m dx dt

)tkx(iAe)t,x( ( )

22 ( ) 2 ( )

2

( )

( )

( )

i kx t

i kx t i kx t

i kx t

ik Aex

ik Ae k Aex

i Aet

Check it. Take the derivatives:

2 2

2km

So, it works if:

That’s the same as:2

2p Em

Lecture 13, p 12

Time-dependence of Superpositions A particle can be in a superposition of states that have different energies.

This superposition is still a solution of the time-dependent SEQ, butnot of the time-independent SEQ, because two different E’s are involved.

Two questions: How does this superposition evolve with time?

For example, is the probability density still stationary? What happens if we measure the particle’s energy?

First, look at the time dependence. Consider the first two energy states in an infinite well. Here’s the superposition:

The two terms have different frequencies, so they oscillate in and out of phase.

U= U=(x)

0 L x

1 21 2( , ) ( ) ( ) i t i tx t x e x e

22 2 1, 4

E E E

21

1 1 2, 8

E hEmL

falstad.com/qm1d/

Lecture 13, p 13

Particle Motion in a Well

The frequency of oscillation is = 2-1 (E2-E1)/, or f = (E2-E1)/h. This is precisely the frequency of a photon that would make a transition between the two states.

So, |(x,t)|2 oscillates between:

i -ie e 2cos We used the identity:

The probability density is given by: |(x,t)|2 :

1 2 21 2 12 2 2 2 cos(( )t)(x, t)

Interference term

2 21 2( , ) ( )x t

Particle localized on left side of well:

(x,t+)

0 xL

Probability

In phase: (cos = +1)

Particle localized on right side of well:

(x,t-)

0 xL

Out of phase: (cos = -1)2 2

1 2( , ) ( )x t

Lecture 13, p 14

Interference BeatsThe motion of the probability density comes from the changing interference between terms in that have different energy.

The beat frequency between two terms is the frequency difference, f2 - f1.

So energy differences are important: E1 - E2 = hf1 - hf2, etc.

Just like in Newtonian physics: absolute energies aren’t important (you can pick where to call U=0 for convenience), only energy differences.

Lecture 13, p 15

Normalizing SuperpositionsWe want the total probability to equal 1, even when the particle is in a superposition of states:

This looks like a mess. However, we’re in luck. Multiply it out:

2 21 2 1totP dx a b dx

2 2 21 2 1 2 1 2 2 1

2 2 2 21 2 1 2 2 1

2 2

* *

* ** *

0 0

a b dx a dx b dx a b dx b a dx

a dx b dx a b dx b a dx

a b

If 1 is normalized If 2 is normalized It is a mathematical theorem thatthese integrals always = 0 if the energies are different.

A normalized superposition must have |a|2 + |b|2 = 1.

= 0.81 + 0.62 is normalized. = 0.51 + 0.52 is not normalized.

Math415

Lecture 13, p 16

Consider a particle in an infinite square well. At t = 0 it is in the state:

with 2(x) and 4(x) both normalized.

1. What is A2? a. 0.5 b. 0.707 c. 0.866

2. At some later time t, what is the probability density at the center of the well?

a. 0 b. 1 c. It depends on the time t.

Act 2

2 2 4( , ) 0.5 ( ) A ( )x t x x

Lecture 13, p 17

Solution

2 2 4( , ) 0.5 ( ) A ( )x t x x

As stated, the question is ambiguous. A2 could be complex. However, let’s assume that A2 is real.

We are told that 2(x) and 4(x) are both normalized. Therefore: 0.52 + |A2|2 = 1 |A2| = sqrt(1 – 0.25) = 0.866

A2 = 0.866 ei

also works, for all .

Consider a particle in an infinite square well. At t = 0 it is in the state:

with 2(x) and 4(x) both normalized.

1. What is A2? a. 0.5 b. 0.707 c. 0.866

2. At some later time t, what is the probability density at the center of the well?

a. 0 b. 1 c. It depends on the time t.

Lecture 13, p 18

Solution

2 2 4( , ) 0.5 ( ) A ( )x t x x

In general, the probability distribution of a superposition of energy eigenstates depends on time. However, 2 and 4 each have a node at L/2.Therefore, every superposition of them also has a node at L/2.

As stated, the question is ambiguous. A2 could be complex. However, let’s assume that A2 is real.

We are told that 2(x) and 4(x) are both normalized. Therefore: 0.52 + |A2|2 = 1 |A2| = sqrt(1 – 0.25) = 0.866

42

2+4

Consider a particle in an infinite square well. At t = 0 it is in the state:

with 2(x) and 4(x) both normalized.

1. What is A2? a. 0.5 b. 0.707 c. 0.866

2. At some later time t, what is the probability density at the center of the well?

a. 0 b. 1 c. It depends on the time t.

A2 = 0.866 ei

also works, for all .

Lecture 13, p 19

Measurements of EnergyWe are now ready to deal with the second question from earlier in the lecture:

What happens when we measure the energy of a particle whose wave function is a superposition of more than one energy state?

If the wave function is in an energy eigenstate (E1, say), then we know with certainty that we will obtain E1 (unless the apparatus is broken).

If the wave function is a superposition ( = a1+b2) of energies E1 and E2, then we aren’t certain what the result will be. However:

We know with certainty that we will only obtain E1 or E2 !!

To be specific, we will never obtain (E1+E2)/2, or any other value.

What about a and b?

|a|2 and |b|2 are the probabilities of obtaining E1 and E2, respectively.

That’s why we normalize the wave function to make |a|2 + |b|2 =1.

We can’t provethis statement. It is one of the fundamental postulates of quantum theory.Treat it as an empirical fact.

Lecture 13, p 20

Consider a particle in an infinite well.It is in the state:

with 2(x) and 4(x) both normalized.

We now measure the energy of the particle. What value is obtained?

a. E2 b. E4 c. 0.25 E2 + 0.75 E4 d. It depends on when we measure the energy.

Not part of this act, but an important question, nevertheless:

If E2 is observed, what is the state of the particle after the measurement?

Act 3

2 4( , ) 0.5 ( , ) 0.866 ( , ) x t x t x t

Lecture 13, p 21

Consider a particle in an infinite well.It is in the state:

with 2 and 4 both normalized.

We now measure the energy of the particle. What value is obtained?

a. E2 b. E4 c. 0.25 E2 + 0.75 E4 d. It depends on when we measure the energy.

Not part of this act, but an important question, nevertheless:

If E2 is observed, what is the state of the particle after the measurement?

Solution

We can only get one of the eigenvalues, E2 or E4. (not answer c)The probability of measuring E2 is 25%.The probability of measuring E4 is 75%.Note: depends on time, but 0.5 and 0.866 don’t. So, d is not correct.

The average energy (if we were to measure a large number of similar particles) is the weighted sum of the energies: 0.25 E2 + 0.75 E4.

2 4( , ) 0.5 ( , ) 0.866 ( , )x t x t x t

Lecture 13, p 22

If E2 is observed, what is the state of the particle after the measurement?

We start out with this wave function:

Before we make the measurement, we can’t predict the result of an energy measurement with certainty.

However, after the measurement, we know with certainty that E = E2. To be specific: We know that a second measurement will yield E2. (Why?)

Therefore, after obtaining E2, the wave function must now be:

That is, the wave function has “collapsed” to the state that corresponds with the result we obtained.

This is one of the weirder features of QM, and is the principal reason that Einstein never accepted QM as a complete theory.

(“God does not place dice!”)

The “Collapse” of the Wave Function

2 4( , ) 0.5 ( , ) 0.866 ( , )x t x t x t

2( , ) ( , ) x t x t

Lecture 13, p 23

Supplement:Time-Energy Uncertainty

PrincipleNow that we are considering time-dependent problems, it is a good time to introduce another application of the Heisenberg Uncertainty Principle, based on measurements of energy and time. We start from our previous result:

Sometimes this is further transformed as follows:

The last line is a standard result from Fourier wave analysis; this should not surprise us – the Uncertainty Principle arises simply because particles behave as waves that are oscillating in time as well as in space. xp x c p E tc

1 1/ 2

E t t

t f t

Lecture 13, p 24

14

15

1 2 11/ 2

1/(2 0.4 10 Hz)

s 44 fs10

t f tt f

A particular optical fiber transmits light over the range 1300-1600 nm (corresponding to a frequency range of 2.3x1014 Hz to 1.9x1014 Hz). How long (approximately) is the shortest pulse that can propagate down this fiber?

E t Uncertainty Principle Example*

Note: This means the upper limit to data transmission is ~1/(4fs) = 2.5x1014 bits/second = 250 Tb/s

*This problem obviously does not require “quantum mechanics” per se. However, due to the Correspondence Principle, QM had better give a consistent result.

Lecture 13, p 25

Next LecturesTunneling of quantum particles Scanning Tunneling Microscope (STM)

Nuclear Decay

Solar Fusion

The Ammonia Maser

Lecture 13, p 26

Supplement:Quantum Information References

Quantum computing Employs superpositions of quantum states - astoundingly good for certain

parallel computation, may use entangled states for error checking www.newscientist.com/nsplus/insight/quantum/48.html http://www.cs.caltech.edu/~westside/quantum-intro.html

Quantum cryptography Employs single photons or entangled pairs of photons to generate a secret key.

Can determine if there has been eavesdropping in information transfer cam.qubit.org/articles/crypto/quantum.php library.lanl.gov/cgi-bin/getfile?00783355.pdf

Quantum teleportation Employs an entangled state to produce an exact replica of a third quantum state at a different

point in space www.research.ibm.com/quantuminfo/teleportation www.quantum.univie.ac.at/research/photonentangle/teleport/index.html Scientific American, April 2000

Supplement: Free particle Supplement: Free particle motionmotion

It turns out (next slide) that the constructive interference region for a matter wavepacket moves at the “group velocity”

v = h/m = p/m

So there’s a simple correspondence between the quantum picture and our classical picture of particles moving around with momentum p = mv.

But the quantum packet will spread out in the long run, since it has a range of p, so the correspondence is never perfect.

Position of maximum probability moves and the width of probability distribution spreads out

x x x

Supplement: Group velocitySupplement: Group velocity Say a wave-packet starts out at x=0 at t=0.

meaning each harmonic component has the same phase there. After time t

the harmonic component at w1 will have changed phase by t1 the harmonic component at w2 will have changed phase by t2 The phase difference between these components at x=0

will now be t(2 - 1) To find the point x where they’re in phase, we need to find where the phase difference from moving downstream by x cancels that: t(2 - 1)=x(k2-k1) Or for small differences in ,k: td=xdk

Result

vg=x/t=d/dk

In this case vg=p/m